# Introduction to Disjoint Set (Union-Find Algorithm)

## What is a Disjoint set data structure?

Two sets are called disjoint sets if they don’t have any element in common, the intersection of sets is a null set.

A data structure that stores non overlapping or disjoint subset of elements is called disjoint set data structure. The disjoint set data structure supports following operations:

• Adding new sets to the disjoint set.
• Merging disjoint sets to a single disjoint set using Union operation.
• Finding representative of a disjoint set using Find operation.
• Check if two sets are disjoint or not.

Consider a situation with a number of persons and the following tasks to be performed on them:

• Add a new friendship relation, i.e. a person x becomes the friend of another person y i.e adding new element to a set.
• Find whether individual x is a friend of individual y (direct or indirect friend)

Examples:

We are given 10 individuals say, a, b, c, d, e, f, g, h, i, j

Following are relationships to be added:
a <-> b
b <-> d
c <-> f
c <-> i
j <-> e
g <-> j

Given queries like whether a is a friend of d or not. We basically need to create following 4 groups and maintain a quickly accessible connection among group items:
G1 = {a, b, d}
G2 = {c, f, i}
G3 = {e, g, j}
G4 = {h}

### Find whether x and y belong to the same group or not, i.e. to find if x and y are direct/indirect friends.

Partitioning the individuals into different sets according to the groups in which they fall. This method is known as a Disjoint set Union which maintains a collection of Disjoint sets and each set is represented by one of its members.

To answer the above question two key points to be considered are:

• How to Resolve sets? Initially, all elements belong to different sets. After working on the given relations, we select a member as a representative. There can be many ways to select a representative, a simple one is to select with the biggest index.
• Check if 2 persons are in the same group? If representatives of two individuals are the same, then they’ll become friends.

### Data Structures used are:

Array: An array of integers is called Parent[]. If we are dealing with N items, i’th element of the array represents the i’th item. More precisely, the i’th element of the Parent[] array is the parent of the i’th item. These relationships create one or more virtual trees.

Tree: It is a Disjoint set. If two elements are in the same tree, then they are in the same Disjoint set. The root node (or the topmost node) of each tree is called the representative of the set. There is always a single unique representative of each set. A simple rule to identify a representative is if ‘i’ is the representative of a set, then Parent[i] = i. If i is not the representative of his set, then it can be found by traveling up the tree until we find the representative.

## Operations on Disjoint Set Data Structures:

1. Find
2. Union

### 1. Find:

Can be implemented by recursively traversing the parent array until we hit a node that is the parent of itself.

## C++

 // Finds the representative of the set// that i is an element of #includeusing namespace std; int find(int i) {     // If i is the parent of itself    if (parent[i] == i) {         // Then i is the representative of        // this set        return i;    }    else {         // Else if i is not the parent of        // itself, then i is not the        // representative of his set. So we        // recursively call Find on its parent        return find(parent[i]);    }} // The code is contributed by Nidhi goel

## Java

 // Finds the representative of the set// that i is an element ofimport java.io.*; class GFG {     static int find(int i)     {         // If i is the parent of itself        if (parent[i] == i) {             // Then i is the representative of            // this set            return i;        }        else {             // Else if i is not the parent of            // itself, then i is not the            // representative of his set. So we            // recursively call Find on its parent            return find(parent[i]);        }    }} // The code is contributed by Nidhi goel

## Python3

 # Finds the representative of the set# that i is an element of def find(i):     # If i is the parent of itself    if (parent[i] == i):         # Then i is the representative of        # this set        return i    else:         # Else if i is not the parent of        # itself, then i is not the        # representative of his set. So we        # recursively call Find on its parent        return find(parent[i])  # The code is contributed by Nidhi goel

## C#

 using System; public class GFG{     // Finds the representative of the set    // that i is an element of    public static int find(int i)    {         // If i is the parent of itself        if (parent[i] == i) {             // Then i is the representative of            // this set            return i;        }        else {             // Else if i is not the parent of            // itself, then i is not the            // representative of his set. So we            // recursively call Find on its parent            return find(parent[i]);        }    }}

## Javascript



Time complexity: This approach is inefficient and can take O(n) time in worst case.

### 2. Union:

It takes two elements as input and finds the representatives of their sets using the Find operation, and finally puts either one of the trees (representing the set) under the root node of the other tree.

## C++

 // Unites the set that includes i// and the set that includes j #include using namespace std; void union(int i, int j) {     // Find the representatives    // (or the root nodes) for the set    // that includes i    int irep = this.Find(i),     // And do the same for the set    // that includes j    int jrep = this.Find(j);     // Make the parent of iâ€™s representative    // be jâ€™s  representative effectively    // moving all of iâ€™s set into jâ€™s set)    this.Parent[irep] = jrep;}

## Java

 import java.util.Arrays; public class UnionFind {    private int[] parent;     public UnionFind(int size) {        // Initialize the parent array with each element as its own representative        parent = new int[size];        for (int i = 0; i < size; i++) {            parent[i] = i;        }    }     // Find the representative (root) of the set that includes element i    public int find(int i) {        if (parent[i] == i) {            return i; // i is the representative of its own set        }        // Recursively find the representative of the parent until reaching the root        parent[i] = find(parent[i]); // Path compression        return parent[i];    }     // Unite (merge) the set that includes element i and the set that includes element j    public void union(int i, int j) {        int irep = find(i); // Find the representative of set containing i        int jrep = find(j); // Find the representative of set containing j         // Make the representative of i's set be the representative of j's set        parent[irep] = jrep;    }     public static void main(String[] args) {        int size = 5; // Replace with your desired size        UnionFind uf = new UnionFind(size);         // Perform union operations as needed        uf.union(1, 2);        uf.union(3, 4);         // Check if elements are in the same set        boolean inSameSet = uf.find(1) == uf.find(2);        System.out.println("Are 1 and 2 in the same set? " + inSameSet);    }}

## Python3

 # Unites the set that includes i# and the set that includes j def union(parent, rank, i, j):    # Find the representatives    # (or the root nodes) for the set    # that includes i    irep = find(parent, i)         # And do the same for the set    # that includes j    jrep = find(parent, j)         # Make the parent of iâ€™s representative    # be jâ€™s  representative effectively    # moving all of iâ€™s set into jâ€™s set)         parent[irep] = jrep

## C#

 using System; public class UnionFind{    private int[] parent;     public UnionFind(int size)    {        // Initialize the parent array with each element as its own representative        parent = new int[size];        for (int i = 0; i < size; i++)        {            parent[i] = i;        }    }     // Find the representative (root) of the set that includes element i    public int Find(int i)    {        if (parent[i] == i)        {            return i; // i is the representative of its own set        }        // Recursively find the representative of the parent until reaching the root        parent[i] = Find(parent[i]); // Path compression        return parent[i];    }     // Unite (merge) the set that includes element i and the set that includes element j    public void Union(int i, int j)    {        int irep = Find(i); // Find the representative of set containing i        int jrep = Find(j); // Find the representative of set containing j         // Make the representative of i's set be the representative of j's set        parent[irep] = jrep;    }     public static void Main()    {        int size = 5; // Replace with your desired size        UnionFind uf = new UnionFind(size);         // Perform union operations as needed        uf.Union(1, 2);        uf.Union(3, 4);         // Check if elements are in the same set        bool inSameSet = uf.Find(1) == uf.Find(2);        Console.WriteLine("Are 1 and 2 in the same set? " + inSameSet);    }}

## Javascript

 // JavaScript code for the approach // Unites the set that includes i// and the set that includes jfunction union(parent, rank, i, j) { // Find the representatives// (or the root nodes) for the set// that includes ilet irep = find(parent, i); // And do the same for the set// that includes jlet jrep = find(parent, j); // Make the parent of iâ€™s representative// be jâ€™s representative effectively// moving all of iâ€™s set into jâ€™s set) parent[irep] = jrep;}

Time complexity: This approach is inefficient and could lead to tree of length O(n) in worst case.

### Optimizations (Union by Rank/Size and Path Compression):

The efficiency depends heavily on which tree get attached to the other. There are 2 ways in which it can be done. First is Union by Rank, which considers height of the tree as the factor and Second is Union by Size, which considers size of the tree as the factor while attaching one tree to the other . This method along with Path Compression gives complexity of nearly constant time.

### Path Compression (Modifications to Find()):

It speeds up the data structure by compressing the height of the trees. It can be achieved by inserting a small caching mechanism into the Find operation. Take a look at the code for more details:

## C++

 // Finds the representative of the set that i// is an element of. #include using namespace std; int find(int i) {     // If i is the parent of itself    if (Parent[i] == i) {         // Then i is the representative         return i;    }    else {          // Recursively find the representative.        int result = find(Parent[i]);         // We cache the result by moving iâ€™s node         // directly under the representative of this        // set        Parent[i] = result;               // And then we return the result        return result;     }}

## Java

 // Finds the representative of the set that i// is an element of.import java.io.*;import java.util.*; static int find(int i) {     // If i is the parent of itself    if (Parent[i] == i) {         // Then i is the representative         return i;    }    else {          // Recursively find the representative.        int result = find(Parent[i]);         // We cache the result by moving iâ€™s node         // directly under the representative of this        // set        Parent[i] = result;               // And then we return the result        return result;     }} // The code  is contributed by Arushi jindal.

## Python3

 #  Finds the representative of the set that i# is an element of.  def find(i):     # If i is the parent of itself    if Parent[i] == i:         # Then i is the representative         return i    else:         # Recursively find the representative.        result = find(Parent[i])         # We cache the result by moving iâ€™s node         # directly under the representative of this        # set        Parent[i] = result               # And then we return the result        return result # The code is contributed by Arushi  Jindal.

## C#

 using System; // Finds the representative of the set that i// is an element of.public static int find(int i) {     // If i is the parent of itself    if (Parent[i] == i) {         // Then i is the representative         return i;    }    else {          // Recursively find the representative.        int result = find(Parent[i]);         // We cache the result by moving iâ€™s node         // directly under the representative of this        // set        Parent[i] = result;               // And then we return the result        return result;     }} // The code is contributed by Arushi Jindal.

## Javascript

 // Finds the representative of the set that i// is an element of.  function find(i) {     // If i is the parent of itself    if (Parent[i] == i) {         // Then i is the representative         return i;    }    else {          // Recursively find the representative.        let result = find(Parent[i]);         // We cache the result by moving iâ€™s node         // directly under the representative of this        // set        Parent[i] = result;               // And then we return the result        return result;     }} // The code is contributed by Arushi  Jindal.

Time Complexity: O(log n) on average per call.

### Union by Rank:

First of all, we need a new array of integers called rank[]. The size of this array is the same as the parent array Parent[]. If i is a representative of a set, rank[i] is the height of the tree representing the set.
Now recall that in the Union operation, it doesnâ€™t matter which of the two trees is moved under the other. Now what we want to do is minimize the height of the resulting tree. If we are uniting two trees (or sets), letâ€™s call them left and right, then it all depends on the rank of left and the rank of right

• If the rank of left is less than the rank of right, then itâ€™s best to move left under right, because that wonâ€™t change the rank of right (while moving right under left would increase the height). In the same way, if the rank of right is less than the rank of left, then we should move right under left.
• If the ranks are equal, it doesnâ€™t matter which tree goes under the other, but the rank of the result will always be one greater than the rank of the trees.

## C++

 // Unites the set that includes i and the set// that includes j by rank #include using namespace std; void unionbyrank(int i, int j) {     // Find the representatives (or the root nodes)    // for the set that includes i    int irep = this.find(i);     // And do the same for the set that includes j    int jrep = this.Find(j);     // Elements are in same set, no need to    // unite anything.    if (irep == jrep)        return;           // Get the rank of iâ€™s tree    irank = Rank[irep],     // Get the rank of jâ€™s tree    jrank = Rank[jrep];     // If iâ€™s rank is less than jâ€™s rank    if (irank < jrank) {         // Then move i under j        this.parent[irep] = jrep;    }     // Else if jâ€™s rank is less than iâ€™s rank    else if (jrank < irank) {         // Then move j under i        this.Parent[jrep] = irep;    }     // Else if their ranks are the same    else {         // Then move i under j (doesnâ€™t matter        // which one goes where)        this.Parent[irep] = jrep;         // And increment the result treeâ€™s        // rank by 1        Rank[jrep]++;    }}

## Java

 public class DisjointSet {     private int[] parent;    private int[] rank;     // Constructor to initialize the DisjointSet data    // structure    public DisjointSet(int size)    {        parent = new int[size];        rank = new int[size];         // Initialize each element as a separate set with        // rank 0        for (int i = 0; i < size; i++) {            parent[i] = i;            rank[i] = 0;        }    }     // Function to find the representative (or the root    // node) of a set with path compression    private int find(int i)    {        if (parent[i] != i) {            parent[i] = find(parent[i]); // Path compression        }        return parent[i];    }     // Unites the set that includes i and the set that    // includes j by rank    public void unionByRank(int i, int j)    {        // Find the representatives (or the root nodes) for        // the set that includes i and j        int irep = find(i);        int jrep = find(j);         // Elements are in the same set, no need to unite        // anything        if (irep == jrep) {            return;        }         // Get the rank of i's tree        int irank = rank[irep];         // Get the rank of j's tree        int jrank = rank[jrep];         // If i's rank is less than j's rank        if (irank < jrank) {            // Move i under j            parent[irep] = jrep;        }        // Else if j's rank is less than i's rank        else if (jrank < irank) {            // Move j under i            parent[jrep] = irep;        }        // Else if their ranks are the same        else {            // Move i under j (doesn't matter which one goes            // where)            parent[irep] = jrep;            // Increment the result tree's rank by 1            rank[jrep]++;        }    }     // Example usage    public static void main(String[] args)    {        int size = 5;        DisjointSet ds = new DisjointSet(size);         // Perform some union operations        ds.unionByRank(0, 1);        ds.unionByRank(2, 3);        ds.unionByRank(1, 3);         // Find the representative of each element and print        // the result        for (int i = 0; i < size; i++) {            System.out.println(                "Element " + i                + " belongs to the set with representative "                + ds.find(i));        }    }}

## Python3

 class DisjointSet:    def __init__(self, size):        self.parent = [i for i in range(size)]        self.rank = [0] * size     # Function to find the representative (or the root node) of a set    def find(self, i):        # If i is not the representative of its set, recursively find the representative        if self.parent[i] != i:            self.parent[i] = self.find(self.parent[i])  # Path compression        return self.parent[i]     # Unites the set that includes i and the set that includes j by rank    def union_by_rank(self, i, j):        # Find the representatives (or the root nodes) for the set that includes i and j        irep = self.find(i)        jrep = self.find(j)         # Elements are in the same set, no need to unite anything        if irep == jrep:            return         # Get the rank of i's tree        irank = self.rank[irep]         # Get the rank of j's tree        jrank = self.rank[jrep]         # If i's rank is less than j's rank        if irank < jrank:            # Move i under j            self.parent[irep] = jrep        # Else if j's rank is less than i's rank        elif jrank < irank:            # Move j under i            self.parent[jrep] = irep        # Else if their ranks are the same        else:            # Move i under j (doesn't matter which one goes where)            self.parent[irep] = jrep            # Increment the result tree's rank by 1            self.rank[jrep] += 1     def main(self):        # Example usage        size = 5        ds = DisjointSet(size)         # Perform some union operations        ds.union_by_rank(0, 1)        ds.union_by_rank(2, 3)        ds.union_by_rank(1, 3)         # Find the representative of each element        for i in range(size):            print(f"Element {i} belongs to the set with representative {ds.find(i)}")  # Creating an instance and calling the main methodds = DisjointSet(size=5)ds.main()

## C#

 using System; class DisjointSet {    private int[] parent;    private int[] rank;     public DisjointSet(int size) {        parent = new int[size];        rank = new int[size];         // Initialize each element as a separate set        for (int i = 0; i < size; i++) {            parent[i] = i;            rank[i] = 0;        }    }     // Function to find the representative (or the root node) of a set    private int Find(int i) {        // If i is not the representative of its set, recursively find the representative        if (parent[i] != i) {            parent[i] = Find(parent[i]); // Path compression        }        return parent[i];    }     // Unites the set that includes i and the set that includes j by rank    public void UnionByRank(int i, int j) {        // Find the representatives (or the root nodes) for the set that includes i and j        int irep = Find(i);        int jrep = Find(j);         // Elements are in the same set, no need to unite anything        if (irep == jrep) {            return;        }         // Get the rank of i's tree        int irank = rank[irep];         // Get the rank of j's tree        int jrank = rank[jrep];         // If i's rank is less than j's rank        if (irank < jrank) {            // Move i under j            parent[irep] = jrep;        }        // Else if j's rank is less than i's rank        else if (jrank < irank) {            // Move j under i            parent[jrep] = irep;        }        // Else if their ranks are the same        else {            // Move i under j (doesn't matter which one goes where)            parent[irep] = jrep;            // Increment the result tree's rank by 1            rank[jrep]++;        }    }     static void Main() {        // Example usage        int size = 5;        DisjointSet ds = new DisjointSet(size);         // Perform some union operations        ds.UnionByRank(0, 1);        ds.UnionByRank(2, 3);        ds.UnionByRank(1, 3);         // Find the representative of each element        for (int i = 0; i < size; i++) {            Console.WriteLine("Element " + i + " belongs to the set with representative " + ds.Find(i));        }    }}

## Javascript

 // JavaScript Program for the above approachunionbyrank(i, j) {let irep = this.find(i); // Find representative of set including ilet jrep = this.find(j); // Find representative of set including j if (irep === jrep) {return; // Elements are already in the same set} let irank = this.rank[irep]; // Rank of set including ilet jrank = this.rank[jrep]; // Rank of set including j if (irank < jrank) {this.parent[irep] = jrep; // Make j's representative parent of i's representative} else if (jrank < irank) {this.parent[jrep] = irep; // Make i's representative parent of j's representative} else {this.parent[irep] = jrep; // Make j's representative parent of i's representativethis.rank[jrep]++;        // Increment the rank of the resulting set}

### Union by Size:

Again, we need a new array of integers called size[]. The size of this array is the same as the parent array Parent[]. If i is a representative of a set, size[i] is the number of the elements in the tree representing the set.
Now we are uniting two trees (or sets), letâ€™s call them left and right, then in this case it all depends on the size of left and the size of right tree (or set).

• If the size of left is less than the size of right, then itâ€™s best to move left under right and increase size of right by size of left. In the same way, if the size of right is less than the size of left, then we should move right under left. and increase size of left by size of right.
• If the sizes are equal, it doesnâ€™t matter which tree goes under the other.

## C++

 // Unites the set that includes i and the set// that includes j by size #include using namespace std; void unionBySize(int i, int j) {     // Find the representatives (or the root nodes)    // for the set that includes i    int irep = find(i);     // And do the same for the set that includes j    int jrep = find(j);     // Elements are in the same set, no need to    // unite anything.    if (irep == jrep)        return;     // Get the size of iâ€™s tree    int isize = Size[irep];     // Get the size of jâ€™s tree    int jsize = Size[jrep];     // If iâ€™s size is less than jâ€™s size    if (isize < jsize) {         // Then move i under j        Parent[irep] = jrep;         // Increment j's size by i's size        Size[jrep] += Size[irep];    }     // Else if jâ€™s size is less than iâ€™s size    else {         // Then move j under i        Parent[jrep] = irep;         // Increment i's size by j's size        Size[irep] += Size[jrep];    }}

## Java

 // Java program for the above approachimport java.util.Arrays; class UnionFind {     private int[] Parent;    private int[] Size;     public UnionFind(int n)    {        // Initialize Parent array        Parent = new int[n];        for (int i = 0; i < n; i++) {            Parent[i] = i;        }         // Initialize Size array with 1s        Size = new int[n];        Arrays.fill(Size, 1);    }     // Function to find the representative (or the root    // node) for the set that includes i    public int find(int i)    {        if (Parent[i] != i) {            // Path compression: Make the parent of i the            // root of the set            Parent[i] = find(Parent[i]);        }        return Parent[i];    }     // Unites the set that includes i and the set that    // includes j by size    public void unionBySize(int i, int j)    {        // Find the representatives (or the root nodes) for        // the set that includes i        int irep = find(i);         // And do the same for the set that includes j        int jrep = find(j);         // Elements are in the same set, no need to unite        // anything.        if (irep == jrep)            return;         // Get the size of iâ€™s tree        int isize = Size[irep];         // Get the size of jâ€™s tree        int jsize = Size[jrep];         // If iâ€™s size is less than jâ€™s size        if (isize < jsize) {            // Then move i under j            Parent[irep] = jrep;             // Increment j's size by i's size            Size[jrep] += Size[irep];        }        // Else if jâ€™s size is less than iâ€™s size        else {            // Then move j under i            Parent[jrep] = irep;             // Increment i's size by j's size            Size[irep] += Size[jrep];        }    }} public class GFG {     public static void main(String[] args)    {        // Example usage        int n = 5;        UnionFind unionFind = new UnionFind(n);         // Perform union operations        unionFind.unionBySize(0, 1);        unionFind.unionBySize(2, 3);        unionFind.unionBySize(0, 4);         // Print the representative of each element after        // unions        for (int i = 0; i < n; i++) {            System.out.println("Element " + i                               + ": Representative = "                               + unionFind.find(i));        }    }} // This code is contributed by Susobhan Akhuli

## Python3

 # Python program for the above approachclass UnionFind:    def __init__(self, n):        # Initialize Parent array        self.Parent = list(range(n))         # Initialize Size array with 1s        self.Size = [1] * n     # Function to find the representative (or the root node) for the set that includes i    def find(self, i):        if self.Parent[i] != i:            # Path compression: Make the parent of i the root of the set            self.Parent[i] = self.find(self.Parent[i])        return self.Parent[i]     # Unites the set that includes i and the set that includes j by size    def unionBySize(self, i, j):        # Find the representatives (or the root nodes) for the set that includes i        irep = self.find(i)         # And do the same for the set that includes j        jrep = self.find(j)         # Elements are in the same set, no need to unite anything.        if irep == jrep:            return         # Get the size of iâ€™s tree        isize = self.Size[irep]         # Get the size of jâ€™s tree        jsize = self.Size[jrep]         # If iâ€™s size is less than jâ€™s size        if isize < jsize:            # Then move i under j            self.Parent[irep] = jrep             # Increment j's size by i's size            self.Size[jrep] += self.Size[irep]        # Else if jâ€™s size is less than iâ€™s size        else:            # Then move j under i            self.Parent[jrep] = irep             # Increment i's size by j's size            self.Size[irep] += self.Size[jrep] # Example usagen = 5unionFind = UnionFind(n) # Perform union operationsunionFind.unionBySize(0, 1)unionFind.unionBySize(2, 3)unionFind.unionBySize(0, 4) # Print the representative of each element after unionsfor i in range(n):    print("Element {}: Representative = {}".format(i, unionFind.find(i))) # This code is contributed by Susobhan Akhuli

## C#

 using System; class UnionFind{    private int[] Parent;    private int[] Size;     public UnionFind(int n)    {        // Initialize Parent array        Parent = new int[n];        for (int i = 0; i < n; i++)        {            Parent[i] = i;        }         // Initialize Size array with 1s        Size = new int[n];        for (int i = 0; i < n; i++)        {            Size[i] = 1;        }    }     // Function to find the representative (or the root node) for the set that includes i    public int Find(int i)    {        if (Parent[i] != i)        {            // Path compression: Make the parent of i the root of the set            Parent[i] = Find(Parent[i]);        }        return Parent[i];    }     // Unites the set that includes i and the set that includes j by size    public void UnionBySize(int i, int j)    {        // Find the representatives (or the root nodes) for the set that includes i        int irep = Find(i);         // And do the same for the set that includes j        int jrep = Find(j);         // Elements are in the same set, no need to unite anything.        if (irep == jrep)            return;         // Get the size of iâ€™s tree        int isize = Size[irep];         // Get the size of jâ€™s tree        int jsize = Size[jrep];         // If iâ€™s size is less than jâ€™s size        if (isize < jsize)        {            // Then move i under j            Parent[irep] = jrep;             // Increment j's size by i's size            Size[jrep] += Size[irep];        }        // Else if jâ€™s size is less than iâ€™s size        else        {            // Then move j under i            Parent[jrep] = irep;             // Increment i's size by j's size            Size[irep] += Size[jrep];        }    }} class Program{    static void Main()    {        // Example usage        int n = 5;        UnionFind unionFind = new UnionFind(n);         // Perform union operations        unionFind.UnionBySize(0, 1);        unionFind.UnionBySize(2, 3);        unionFind.UnionBySize(0, 4);         // Print the representative of each element after unions        for (int i = 0; i < n; i++)        {            Console.WriteLine(\$"Element {i}: Representative = {unionFind.Find(i)}");        }    }}

## Javascript

 unionbysize(i, j) {        let irep = this.find(i); // Find the representative of the set containing i.        let jrep = this.find(j); // Find the representative of the set containing j.         if (irep === jrep) {            return; // Elements are already in the same set.        }         let isize = this.size[irep]; // Size of the set including i.        let jsize = this.size[jrep]; // Size of the set including j.         if (isize < jsize) {            // If i's size is less than j's size, make i's representative             // a child of j's representative.            this.parent[irep] = jrep;            this.size[jrep] += this.size[irep]; // Increment j's size by i's size.        } else {            // If j's size is less than or equal to i's size, make j's representative             // a child of i's representative.            this.parent[jrep] = irep;            this.size[irep] += this.size[jrep]; // Increment i's size by j's size.            if (isize === jsize) {                // If sizes are equal, increment the rank of i's representative.                this.rank[irep]++;            }        }    }

Output
Element 0: Representative = 0
Element 1: Representative = 0
Element 2: Representative = 2
Element 3: Representative = 2
Element 4: Representative = 0

Time complexity: O(log n) without Path Compression.

## C++

 // C++ implementation of disjoint set #include using namespace std; class DisjSet {    int *rank, *parent, n; public:       // Constructor to create and    // initialize sets of n items    DisjSet(int n)    {        rank = new int[n];        parent = new int[n];        this->n = n;        makeSet();    }     // Creates n single item sets    void makeSet()    {        for (int i = 0; i < n; i++) {            parent[i] = i;        }    }     // Finds set of given item x    int find(int x)    {        // Finds the representative of the set        // that x is an element of        if (parent[x] != x) {             // if x is not the parent of itself            // Then x is not the representative of            // his set,            parent[x] = find(parent[x]);             // so we recursively call Find on its parent            // and move i's node directly under the            // representative of this set        }         return parent[x];    }     // Do union of two sets by rank represented    // by x and y.    void Union(int x, int y)    {        // Find current sets of x and y        int xset = find(x);        int yset = find(y);         // If they are already in same set        if (xset == yset)            return;         // Put smaller ranked item under        // bigger ranked item if ranks are        // different        if (rank[xset] < rank[yset]) {            parent[xset] = yset;        }        else if (rank[xset] > rank[yset]) {            parent[yset] = xset;        }         // If ranks are same, then increment        // rank.        else {            parent[yset] = xset;            rank[xset] = rank[xset] + 1;        }    }}; // Driver Codeint main(){         // Function Call    DisjSet obj(5);    obj.Union(0, 2);    obj.Union(4, 2);    obj.Union(3, 1);       if (obj.find(4) == obj.find(0))        cout << "Yes\n";    else        cout << "No\n";    if (obj.find(1) == obj.find(0))        cout << "Yes\n";    else        cout << "No\n";     return 0;}

## Java

 // A Java program to implement Disjoint Set Data// Structure.import java.io.*;import java.util.*; class DisjointUnionSets {    int[] rank, parent;    int n;     // Constructor    public DisjointUnionSets(int n)    {        rank = new int[n];        parent = new int[n];        this.n = n;        makeSet();    }     // Creates n sets with single item in each    void makeSet()    {        for (int i = 0; i < n; i++) {            // Initially, all elements are in            // their own set.            parent[i] = i;        }    }     // Returns representative of x's set    int find(int x)    {        // Finds the representative of the set        // that x is an element of        if (parent[x] != x) {            // if x is not the parent of itself            // Then x is not the representative of            // his set,            parent[x] = find(parent[x]);             // so we recursively call Find on its parent            // and move i's node directly under the            // representative of this set        }         return parent[x];    }     // Unites the set that includes x and the set    // that includes x    void union(int x, int y)    {        // Find representatives of two sets        int xRoot = find(x), yRoot = find(y);         // Elements are in the same set, no need        // to unite anything.        if (xRoot == yRoot)            return;         // If x's rank is less than y's rank        if (rank[xRoot] < rank[yRoot])             // Then move x under y  so that depth            // of tree remains less            parent[xRoot] = yRoot;         // Else if y's rank is less than x's rank        else if (rank[yRoot] < rank[xRoot])             // Then move y under x so that depth of            // tree remains less            parent[yRoot] = xRoot;         else // if ranks are the same        {            // Then move y under x (doesn't matter            // which one goes where)            parent[yRoot] = xRoot;             // And increment the result tree's            // rank by 1            rank[xRoot] = rank[xRoot] + 1;        }    }} // Driver codepublic class Main {    public static void main(String[] args)    {        // Let there be 5 persons with ids as        // 0, 1, 2, 3 and 4        int n = 5;        DisjointUnionSets dus =                 new DisjointUnionSets(n);         // 0 is a friend of 2        dus.union(0, 2);         // 4 is a friend of 2        dus.union(4, 2);         // 3 is a friend of 1        dus.union(3, 1);         // Check if 4 is a friend of 0        if (dus.find(4) == dus.find(0))            System.out.println("Yes");        else            System.out.println("No");         // Check if 1 is a friend of 0        if (dus.find(1) == dus.find(0))            System.out.println("Yes");        else            System.out.println("No");    }}

## Python3

 # Python3 program to implement Disjoint Set Data# Structure. class DisjSet:    def __init__(self, n):        # Constructor to create and        # initialize sets of n items        self.rank = [1] * n        self.parent = [i for i in range(n)]      # Finds set of given item x    def find(self, x):                 # Finds the representative of the set        # that x is an element of        if (self.parent[x] != x):                         # if x is not the parent of itself            # Then x is not the representative of            # its set,            self.parent[x] = self.find(self.parent[x])                         # so we recursively call Find on its parent            # and move i's node directly under the            # representative of this set         return self.parent[x]      # Do union of two sets represented    # by x and y.    def Union(self, x, y):                 # Find current sets of x and y        xset = self.find(x)        yset = self.find(y)         # If they are already in same set        if xset == yset:            return         # Put smaller ranked item under        # bigger ranked item if ranks are        # different        if self.rank[xset] < self.rank[yset]:            self.parent[xset] = yset         elif self.rank[xset] > self.rank[yset]:            self.parent[yset] = xset         # If ranks are same, then move y under        # x (doesn't matter which one goes where)        # and increment rank of x's tree        else:            self.parent[yset] = xset            self.rank[xset] = self.rank[xset] + 1 # Driver codeobj = DisjSet(5)obj.Union(0, 2)obj.Union(4, 2)obj.Union(3, 1)if obj.find(4) == obj.find(0):    print('Yes')else:    print('No')if obj.find(1) == obj.find(0):    print('Yes')else:    print('No') # This code is contributed by ng24_7.

## C#

 // A C# program to implement  // Disjoint Set Data Structure.using System;     class DisjointUnionSets {    int[] rank, parent;    int n;     // Constructor    public DisjointUnionSets(int n)    {        rank = new int[n];        parent = new int[n];        this.n = n;        makeSet();    }     // Creates n sets with single item in each    public void makeSet()    {        for (int i = 0; i < n; i++)        {            // Initially, all elements are in            // their own set.            parent[i] = i;        }    }     // Returns representative of x's set    public int find(int x)    {        // Finds the representative of the set        // that x is an element of        if (parent[x] != x)        {                         // if x is not the parent of itself            // Then x is not the representative of            // his set,            parent[x] = find(parent[x]);             // so we recursively call Find on its parent            // and move i's node directly under the            // representative of this set        }        return parent[x];    }     // Unites the set that includes x and    // the set that includes x    public void union(int x, int y)    {        // Find representatives of two sets        int xRoot = find(x), yRoot = find(y);         // Elements are in the same set,         // no need to unite anything.        if (xRoot == yRoot)            return;         // If x's rank is less than y's rank        if (rank[xRoot] < rank[yRoot])             // Then move x under y so that depth            // of tree remains less            parent[xRoot] = yRoot;         // Else if y's rank is less than x's rank        else if (rank[yRoot] < rank[xRoot])             // Then move y under x so that depth of            // tree remains less            parent[yRoot] = xRoot;         else // if ranks are the same        {            // Then move y under x (doesn't matter            // which one goes where)            parent[yRoot] = xRoot;             // And increment the result tree's            // rank by 1            rank[xRoot] = rank[xRoot] + 1;        }    }} // Driver codeclass GFG {    public static void Main(String[] args)    {        // Let there be 5 persons with ids as        // 0, 1, 2, 3 and 4        int n = 5;        DisjointUnionSets dus =                 new DisjointUnionSets(n);         // 0 is a friend of 2        dus.union(0, 2);         // 4 is a friend of 2        dus.union(4, 2);         // 3 is a friend of 1        dus.union(3, 1);         // Check if 4 is a friend of 0        if (dus.find(4) == dus.find(0))            Console.WriteLine("Yes");        else            Console.WriteLine("No");         // Check if 1 is a friend of 0        if (dus.find(1) == dus.find(0))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by Rajput-Ji

## Javascript

 class DisjSet {    constructor(n) {        this.rank = new Array(n);        this.parent = new Array(n);        this.n = n;        this.makeSet();    }     makeSet() {        for (let i = 0; i < this.n; i++) {            this.parent[i] = i;        }    }     find(x) {        if (this.parent[x] !== x) {            this.parent[x] = this.find(this.parent[x]);        }        return this.parent[x];    }     Union(x, y) {        let xset = this.find(x);        let yset = this.find(y);         if (xset === yset) return;         if (this.rank[xset] < this.rank[yset]) {            this.parent[xset] = yset;        } else if (this.rank[xset] > this.rank[yset]) {            this.parent[yset] = xset;        } else {            this.parent[yset] = xset;            this.rank[xset] = this.rank[xset] + 1;        }    }} // usage examplelet obj = new DisjSet(5);obj.Union(0, 2);obj.Union(4, 2);obj.Union(3, 1); if (obj.find(4) === obj.find(0)) {  console.log("Yes");} else {  console.log("No");}if (obj.find(1) === obj.find(0)) {  console.log("Yes");} else {  console.log("No");}

Output
Yes
No

Time complexity: O(n) for creating n single item sets . The two techniques -path compression with the union by rank/size, the time complexity will reach nearly constant time. It turns out, that the final amortized time complexity is O(Î±(n)), where Î±(n) is the inverse Ackermann function, which grows very steadily (it does not even exceed for n<10600  approximately).

Space complexity: O(n) because we need to store n elements in the Disjoint Set Data Structure.

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